Hi, I'm looking for something with AC male on one end, and AC female
on the other, that can plug an AC device (sump pump) into, that will
emit an audible tone when the device is running and consuming current.
Kind of like an audible pilot light, or an electric meter that
sings.. I want this so I can be aware upstairs when the sump pump runs
in the cellar, since occasionally it sticks on. This would also alert
us to a plumbing problem. It's a dirt cellar.
It seems like would be a simple device, but I don't know what it would
be called.
---
FWB
+----+
+-------------|~ +|------+-----+
| +-----|~ -|--+ |+ |
| | +----+ | [C] [BUZZER]
MAINS>---+--[R]--+----+ | | |
| +---+-----+
[PUMP]
|
MAINS>----------------+
Select the lowest voltage buzzer (beeper, Sonalert, etc.) you can
find, and choose R so that the current drawn by the pump motor will
drop the voltage needed by the buzzer plus 1.4V for the diodes in the
bridge. That is:
R = (Vbuzzer + 1.4V) / I motor
The capacitor, C, will be:
C = IdT/dV
where C = the capacitance in Farads
I = the steady-state current ion the buzzer
dt = the period of the rectified waveform
dv = the permissible ripple voltage across the buzzer
For example, lets say you have a pump which draws 1A and you've chosen
a 6 volt Sonalert which needs 100mA.
For R,
R = (Vb + 1.4V) / 1A = 7.4/1 = 7.4 ohms.
7.5 ohms is a standard value, and then, since the resistor will drop
7.5V at 1A, it will dissipate 7.5 watts, so you'll need to get
something like a 15 or 20 watt resistor.
A 20 watter won't get as hot as a 10 watter, so that would be the way
to go, IMO. A good idea would be to use a wirewound metal-cased 25
watt resistor which you could bolt onto the inside of a small metal
minibox which you could use to house the rest of the components.
Restricting all of the electrical connections to the inside of the
enclosure and making sure there were no electrical connections made to
the [metal] housing itself would go a long way to reducing the danger
of shock, since you're using the unisolated mains to run the buzzer.
For C, assuming you can tolerate 1/2 a volt of ripple:
C = Idt / dV = 0.1A*0.0083s/0.5V = .00166µF ~ 1700µF
Since the tolerance on the capacitance of cheap electrolytics is
often rated at -20 to +80%, you'd need about 2200µF to make sure you
got 1700µF, and the cap would need to be rated for at least 6V. 12V
wouldn't hurt.
Pretty much any old bridge would work, or you could make up something
out of 1N4000s, so your circuit would wind up looking like this:
FWB
+----+
+----------------|~ +|-------+-------+
| +-----|~ -|--+ |+ |
| | +----+ | [2200] [BUZZER]
MAINS>---+--[7.5R]--+----+ | | |
| +----+-------+
[PUMP]
|
MAINS>-------------------+
If you wanted to cut down on the dissipation you could substitute a
current transformer for the 7.5 ohm resistor, but that's another
story.
Whatever you decide to do be careful, since mains voltage can kill.